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JEE Main 2024 · Physics · Magnetic Effects of Current · medium

A long air-cored solenoid has 2000 turns per metre and cross-sectional area 5 cm^2. A compact coil of 50 turns and resistance 5 Ω is wound tightly…

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A long air-cored solenoid has 20002000 turns per metre and cross-sectional area 5 cm25\ \mathrm{cm^2}. A compact coil of 5050 turns and resistance 5 Ω5\ \Omega is wound tightly round its middle. The solenoid current is reversed from +4 A+4\ \mathrm{A} to −4 A-4\ \mathrm{A}. The total charge that flows through the coil is
  1. A. 32π μC32\pi\ \mu\mathrm{C}
  2. B. 16π μC16\pi\ \mu\mathrm{C}
  3. C. 64π μC64\pi\ \mu\mathrm{C}
  4. D. 8π μC8\pi\ \mu\mathrm{C}
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Answer: (A) 32π μC32\pi\ \mu\mathrm{C}

Field inside the solenoid B=μ0nIB=\mu_0nI, so the flux linked with the coil is Nμ0nIAN\mu_0nIA. Change in flux linkage on reversal: Δ(NΦ)=Nμ0nA (2×4)=50×4π×10−7×2000×5×10−4×8=1.6π×10−4 Wb\Delta(N\Phi)=N\mu_0nA\,(2\times4)=50\times4\pi\times10^{-7}\times2000\times5\times10^{-4}\times8=1.6\pi\times10^{-4}\ \mathrm{Wb}. Charge q=Δ(NΦ)R=1.6π×10−45=3.2π×10−5 C=32π μCq=\dfrac{\Delta(N\Phi)}{R}=\dfrac{1.6\pi\times10^{-4}}{5}=3.2\pi\times10^{-5}\ \mathrm{C}=32\pi\ \mu\mathrm{C}. Taking the change as 4 A4\ \mathrm{A} instead of 8 A8\ \mathrm{A} gives the trap 16π μC16\pi\ \mu\mathrm{C}.
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